Hdu6106 Classes（2017多校第6场）

最新推荐文章于 2017-08-14 19:27:36 发布

Jelly_acm

最新推荐文章于 2017-08-14 19:27:36 发布

阅读量627

点赞数

分类专栏： HDU

本文链接：https://blog.youkuaiyun.com/Jelly_acm/article/details/77074263

版权

HDU 专栏收录该内容

181 篇文章

订阅专栏

Classes

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 244 Accepted Submission(s): 158

Problem Description

The school set up three elective courses, assuming that these courses are A, B, C. N classes of students enrolled in these courses.
Now the school wants to count the number of students who enrolled in at least one course in each class and records the maximum number of students.
Each class uploaded 7 data, the number of students enrolled in course A in the class, the number of students enrolled in course B, the number of students enrolled in course C, the number of students enrolled in course AB, the number of students enrolled in course BC, the number of students enrolled in course AC, the number of students enrolled in course ABC. The school can calculate the number of students in this class based on these 7 data.
However, due to statistical errors, some data are wrong and these data should be ignored.
Smart you must know how to write a program to find the maximum number of students.

Input

The first line of the input gives the number of test cases T; T test cases follow.
Each case begins with one line with one integer N, indicates the number of class.
Then N lines follow, each line contains 7 data: a, b, c, d, e, f, g, indicates the number of students enrolled in A, B, C, AB, BC, AC, ABC in this class.
It's guaranteed that at least one data is right in each test case.

Limits

T≤100

1≤N≤100

0≤a,b,c,d,e,f,g≤100

Output

For each test case output one line contains one integer denotes the max number of students who enrolled in at least one course among N classes.

Sample Input

      
      
       
       2
2
4 5 4 4 3 2 2
5 3 1 2 0 0 0
2
0 4 10 2 3 4 9
6 12 6 3 5 3 2

Sample Output

      
      
       
       7
15


       
       
        
        
         
         Hint
        
        
In the second test case, the data uploaded by Class 1 is wrong. 
Because we can't find a solution which satisfies the limitation. 
As for Class 2, we can calculate the number of students who only enrolled in course A is 2, 
the number of students who only enrolled in course B is 6, and nobody enrolled in course C,
the number of students who only enrolled in courses A and B is 1, 
the number of students who only enrolled in courses B and C is 3, 
the number of students who only enrolled in courses A and C is 1, 
the number of students who enrolled in all courses  is 2, 
so the total number in Class 2 is 2 + 6 + 0 + 1 + 3 + 1 + 2 = 15.

Source

2017 Multi-University Training Contest - Team 6

——————————————————————————————————

题目的意思是给出一A，B，C，AB，BC，AC，ABC的选课人数，有些可能是伪造的不计算，计算真实的人最多有几个

思路：分别计算出只选A，B，C，只选AB，AC，BC和ABC的人数，判是否都非负

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <set>
#include <string>
#include <cmath>
#include <algorithm>
#include <vector>
#include <bitset>
#include <stack>
#include <queue>
#include <unordered_map>
#include <functional>

using namespace std;

int a[8];
int b[8];
int n;
int main()
{
    int T;
    for(scanf("%d",&T);T--;)
    {
        int sum=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<7;j++)
                scanf("%d",&a[j]);
            b[0] = a[0] - a[3] - a[5] + a[6];
            b[1] = a[1] - a[3] - a[4] + a[6];
            b[2] = a[2] - a[4] - a[5] + a[6];
            b[3] = a[3] - a[6];
            b[4] = a[4] - a[6];
            b[5] = a[5] - a[6];
            b[6] = a[6];
            int ttt = 0;
            for(int j = 0; j < 7; j++)
            {
                if(b[j] < 0)
                {
                    ttt = 0;
                    break;
                }
                ttt += b[j];
            }
            sum = max(sum, ttt);
        }
        printf("%d\n", sum);
    }
    return 0;
}